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It is apparently a result of F. González-Acuña that all closed orientable 3-manifolds contain a fibered knot. (I am not sure exactly where to find a published proof of this result and as an aside I would be interested in hearing about any proofs/references that anyone knows.)

I am wondering to what extent this is true in higher dimensions and specifically in dimension 4. In particular, is there always a fibered 2-sphere in a closed orientable 4-manifold? If not, what about arbitrary genus surfaces?

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    $\begingroup$ For the 3-manifold case, one could use that the figure 8 knot is universal. The pullback of the figure 8 will be a fibered link in the branched cover. One can plumb hopf bands to create a knot then. $\endgroup$
    – Ian Agol
    Jan 21, 2020 at 5:41
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    $\begingroup$ Sorry if this is basic, but what is a fibered surface? $\endgroup$ Jan 21, 2020 at 6:14
  • $\begingroup$ a surface whose complement is fibered $\endgroup$
    – ThiKu
    Feb 3, 2020 at 8:54

2 Answers 2

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I suppose that, for a $n$-manifold $M$, containing a "fibered codimension-2 manifold" $N\subset M$ means that $N$ has trivial normal neighbourhood $\nu N = N\times D^2$ and its complement $M \setminus \nu N$ fibers over $S^1$ with a fibration that extends the projection $\partial (\nu N) = N \times S^1 \to S^1$ of its boundary. The fiber of the fibration is some $(n-1)$-manifold $Y$ with boundary $N$.

Any such manifold $M$ is obtained from a $(n-1)$-manifold with boundary $Y$ and a self-diffeomorphism $\varphi$ of it (that restricts to the identity on the boundary). You use $\varphi$ as the monodromy to construct a fibration over $S^1$ with fiber $Y$, and then cap the boundary component by gluing $\partial Y \times D^2$. Following Gordon, when $\varphi$ is trivial, we may say that $M$ is obtained by spinning $Y$. In this case we get $\pi_1(M) = \pi_1(Y)$.

The first thing to do is to calculate some topological invariants of such a manifold $M$ obtained in this way. We get $\chi(M) = \chi(N)$. So if a four-manifold $M$ contains a fibred 2-sphere it must have $\chi(M)=2$. If it contains a more general fibred connected surface $N$, we get $\chi(M) = \chi(N) \leq 2$. I would also bet that $M$ is even and has zero signature (I haven't checked this). One could look at the possible fundamental groups $\pi_1(M)$ one could get. Geometrisation of 3-manifolds gives some information on the possible $Y$ and the possible $\varphi$.

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  • $\begingroup$ Thank you for this - I did not notice that about the Euler characteristic, and the signature does indeed vanish as @MarcKegel pointed out. How does geometrisation give info about $Y$ and $\phi$? $\endgroup$
    – user101010
    Jan 21, 2020 at 20:16
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A fibered codimension $2$ link in a manifold $M$ can be identified with an open book decomposition of $M$ (the object that is described at the beginning of Bruno Martelli's answer).

The existence of open books was studied in Quinn's paper:

Open book decompositions, and the bordism of automorphisms.

In particular, any odd-dimensional (closed oriented) manifold admits the structure of an open book. In even dimensions there are obstructions. For example, the only surface admitting an open book is $S^2$.

A general obstruction is coming from the signature as Bruno Martelli was pointing out. Given an open book decomposition of $M$, it is relatively easy to construct an orientation reversing diffeomorphisms on the neighborhood of the codimension $2$ manifold and on its complement. Thus the signature has to vanish.

In the above paper, Quinn identifies another (more complicated) algebraic obstruction to the existence of open books in even dimensions larger than $4$ and shows on the other hand that these are the only obstructions.

In dimension $4$ the question about the existence of open book structures seems to be unsolved (apart from that the signature has to vanish). (Quinn's paper uses the $s$-cobordism theorem and thus the argument is probably not working in dimension $4$.)

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